3.441 \(\int \frac {\tan (x)}{(1+5 \tan ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=54 \[ \frac {1}{16 \sqrt {5 \tan ^2(x)+1}}-\frac {1}{12 \left (5 \tan ^2(x)+1\right )^{3/2}}+\frac {1}{32} \tan ^{-1}\left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3670, 444, 51, 63, 203} \[ \frac {1}{32} \tan ^{-1}\left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right )+\frac {1}{16 \sqrt {5 \tan ^2(x)+1}}-\frac {1}{12 \left (5 \tan ^2(x)+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(1 + 5*Tan[x]^2)^(5/2),x]

[Out]

ArcTan[Sqrt[1 + 5*Tan[x]^2]/2]/32 - 1/(12*(1 + 5*Tan[x]^2)^(3/2)) + 1/(16*Sqrt[1 + 5*Tan[x]^2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx &=\operatorname {Subst}\left (\int \frac {x}{\left (1+x^2\right ) \left (1+5 x^2\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1+x) (1+5 x)^{5/2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac {1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{(1+x) (1+5 x)^{3/2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac {1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac {1}{16 \sqrt {1+5 \tan ^2(x)}}+\frac {1}{32} \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {1+5 x}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac {1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac {1}{16 \sqrt {1+5 \tan ^2(x)}}+\frac {1}{80} \operatorname {Subst}\left (\int \frac {1}{\frac {4}{5}+\frac {x^2}{5}} \, dx,x,\sqrt {1+5 \tan ^2(x)}\right )\\ &=\frac {1}{32} \tan ^{-1}\left (\frac {1}{2} \sqrt {1+5 \tan ^2(x)}\right )-\frac {1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac {1}{16 \sqrt {1+5 \tan ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 71, normalized size = 1.31 \[ \frac {(2 \cos (2 x)-3) \sec ^5(x) \left (-6 \cos (x)+8 \cos (3 x)-3 (2 \cos (2 x)-3)^{3/2} \log \left (2 \cos (x)+\sqrt {2 \cos (2 x)-3}\right )\right )}{96 \left (5 \tan ^2(x)+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(1 + 5*Tan[x]^2)^(5/2),x]

[Out]

((-3 + 2*Cos[2*x])*(-6*Cos[x] + 8*Cos[3*x] - 3*(-3 + 2*Cos[2*x])^(3/2)*Log[2*Cos[x] + Sqrt[-3 + 2*Cos[2*x]]])*
Sec[x]^5)/(96*(1 + 5*Tan[x]^2)^(5/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Tan[x]/(1 + 5*Tan[x]^2)^(5/2),x]

[Out]

Could not integrate

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fricas [A]  time = 1.36, size = 76, normalized size = 1.41 \[ \frac {3 \, {\left (25 \, \tan \relax (x)^{4} + 10 \, \tan \relax (x)^{2} + 1\right )} \arctan \left (\frac {5 \, \tan \relax (x)^{2} - 3}{4 \, \sqrt {5 \, \tan \relax (x)^{2} + 1}}\right ) + 4 \, {\left (15 \, \tan \relax (x)^{2} - 1\right )} \sqrt {5 \, \tan \relax (x)^{2} + 1}}{192 \, {\left (25 \, \tan \relax (x)^{4} + 10 \, \tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+5*tan(x)^2)^(5/2),x, algorithm="fricas")

[Out]

1/192*(3*(25*tan(x)^4 + 10*tan(x)^2 + 1)*arctan(1/4*(5*tan(x)^2 - 3)/sqrt(5*tan(x)^2 + 1)) + 4*(15*tan(x)^2 -
1)*sqrt(5*tan(x)^2 + 1))/(25*tan(x)^4 + 10*tan(x)^2 + 1)

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giac [A]  time = 0.61, size = 36, normalized size = 0.67 \[ \frac {15 \, \tan \relax (x)^{2} - 1}{48 \, {\left (5 \, \tan \relax (x)^{2} + 1\right )}^{\frac {3}{2}}} + \frac {1}{32} \, \arctan \left (\frac {1}{2} \, \sqrt {5 \, \tan \relax (x)^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+5*tan(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/48*(15*tan(x)^2 - 1)/(5*tan(x)^2 + 1)^(3/2) + 1/32*arctan(1/2*sqrt(5*tan(x)^2 + 1))

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maple [A]  time = 0.09, size = 41, normalized size = 0.76

method	result	size
derivativedivides	\(\frac {\arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{2}\right )}{32}+\frac {1}{16 \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}-\frac {1}{12 \left (1+5 \left (\tan ^{2}\relax (x )\right )\right )^{\frac {3}{2}}}\)	\(41\)
default	\(\frac {\arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{2}\right )}{32}+\frac {1}{16 \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}-\frac {1}{12 \left (1+5 \left (\tan ^{2}\relax (x )\right )\right )^{\frac {3}{2}}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(1+5*tan(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/32*arctan(1/2*(1+5*tan(x)^2)^(1/2))+1/16/(1+5*tan(x)^2)^(1/2)-1/12/(1+5*tan(x)^2)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \relax (x)}{{\left (5 \, \tan \relax (x)^{2} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+5*tan(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(x)/(5*tan(x)^2 + 1)^(5/2), x)

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mupad [B]  time = 0.51, size = 172, normalized size = 3.19 \[ \frac {\ln \left (\mathrm {tan}\relax (x)-\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}}{5}+\frac {1}{5}{}\mathrm {i}\right )\,1{}\mathrm {i}}{64}+\frac {\ln \left (\mathrm {tan}\relax (x)+\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}}{5}-\frac {1}{5}{}\mathrm {i}\right )\,1{}\mathrm {i}}{64}-\frac {\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,1{}\mathrm {i}}{64}-\frac {\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{64}-\frac {\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}\,1{}\mathrm {i}}{96\,\left (\mathrm {tan}\relax (x)-\frac {\sqrt {5}\,1{}\mathrm {i}}{5}\right )}+\frac {\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}\,1{}\mathrm {i}}{96\,\left (\mathrm {tan}\relax (x)+\frac {\sqrt {5}\,1{}\mathrm {i}}{5}\right )}+\frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}}{240\,\left ({\mathrm {tan}\relax (x)}^2+\frac {2{}\mathrm {i}\,\sqrt {5}\,\mathrm {tan}\relax (x)}{5}-\frac {1}{5}\right )}-\frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}}{240\,\left (-{\mathrm {tan}\relax (x)}^2+\frac {2{}\mathrm {i}\,\sqrt {5}\,\mathrm {tan}\relax (x)}{5}+\frac {1}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(5*tan(x)^2 + 1)^(5/2),x)

[Out]

(log(tan(x) - (2*5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/5 + 1i/5)*1i)/64 + (log(tan(x) + (2*5^(1/2)*(tan(x)^2 + 1/5)^
(1/2))/5 - 1i/5)*1i)/64 - (log(tan(x) - 1i)*1i)/64 - (log(tan(x) + 1i)*1i)/64 - ((tan(x)^2 + 1/5)^(1/2)*1i)/(9
6*(tan(x) - (5^(1/2)*1i)/5)) + ((tan(x)^2 + 1/5)^(1/2)*1i)/(96*(tan(x) + (5^(1/2)*1i)/5)) + (5^(1/2)*(tan(x)^2
+ 1/5)^(1/2))/(240*(tan(x)^2 + (5^(1/2)*tan(x)*2i)/5 - 1/5)) - (5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/(240*((5^(1/2
)*tan(x)*2i)/5 - tan(x)^2 + 1/5))

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sympy [A]  time = 8.11, size = 46, normalized size = 0.85 \[ \frac {\operatorname {atan}{\left (\frac {\sqrt {5 \tan ^{2}{\relax (x )} + 1}}{2} \right )}}{32} + \frac {1}{16 \sqrt {5 \tan ^{2}{\relax (x )} + 1}} - \frac {1}{12 \left (5 \tan ^{2}{\relax (x )} + 1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+5*tan(x)**2)**(5/2),x)

[Out]

atan(sqrt(5*tan(x)**2 + 1)/2)/32 + 1/(16*sqrt(5*tan(x)**2 + 1)) - 1/(12*(5*tan(x)**2 + 1)**(3/2))

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